Show that \(n! \geqslant {2^{n - 1}}\), for \(n \geqslant 1\).

Hence use the comparison test to determine whether the series \(\sum\limits_{n = 1}^\infty  {\frac{1}{{n!}}} \) converges or diverges.

for \(n \geqslant 1,{\text{ }}n! = n(n - 1)(n - 2) \ldots 3 \times 2 \times 1 \geqslant 2 \times 2 \times 2 \ldots 2 \times 2 \times 1 = {2^{n - 1}}\)     M1A1

\( \Rightarrow n! \geqslant {2^{n - 1}}{\text{ for }}n \geqslant 1\)     AG

[2 marks]

\(n! \geqslant {2^{n - 1}} \Rightarrow \frac{1}{{n!}} \leqslant \frac{1}{{{2^{n - 1}}}}{\text{ for }}n \geqslant 1\)     A1

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{2^{n - 1}}}}} \) is a positive converging geometric series     R1

hence \(\sum\limits_{n = 1}^\infty  {\frac{1}{{n!}}} \) converges by the comparison test     R1

[3 marks]

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for n = 1, 2, 3 and therefore true for all n. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n - 1}}}}} \).

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for n = 1, 2, 3 and therefore true for all n. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n - 1}}}}} \).